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What is "negative geometric washout"?
I read that the wing of a hang glider should be tightened with cables so that it has a "negative washout", but I didn't really find anything about what this means

Welcome to avitation.se Entaldalpe! @everyone else: if you have an answer, post it as an answer. Don't send people elsewhere, they came to you for help.

This should answer your question.

I strongly suspect that what you read was an example of poor writing or word choice, or a mis-translation.  Almost certainly what was intended, was to say that the wing should have anhedral-- i.e. when the glider is placed in a position with the mean chord line roughly horizontal, or perhaps with the root chord line horizontal, or the keel tube horizontal (lots of choices here!),1 the leading edges should have some down droop, root to tip.  In flight-- but not when the glider is "unloaded" on the ground2-- this geometry is determined by the length of the lower "side wires".
I've never seen an example of a modern flex-wing hang glider lacking this configuration (no matter which of the three ways listed above we choose to define it).  Nor have I ever seen an example of a modern flex-wing hang glider lacking (normal, positive) washout.
A key question re your question, is "which cables?". The main wing wires ("lower side wires")?  If so, see the above content re anhedral.  The "luff lines"?  If so, then maybe the source is simply in error about which direction of twist would be implied by (normal, positive) washout versus this so-called "negative washout".  The glider needs to have positive washout, and the "luff lines" do have some influence on that,3 at least when the glider is fully unloaded on the ground.
"Washout" means that the chord line of the wing tip airfoil is inclined in a leading-edge-down, trailing-edge-up manner in relation to the chord line of the wing root airfoil.  Like if you grabbed hold of the wing tip and twisted it so the trailing edge was raised.  (Note that the aerodynamic loads tend to do this naturally.)  A hang glider wing that was twisted in the opposite direction ("negative washout"), while actually loaded in actual flight, would be a death trap.
"Geometric" usually means that we are simply measuring things with a ruler or protractor, without knowing the true direction or the airflow at any point on the aircraft, and therefore not accounting for "twists" or changes in the direction of the airflow over the glider's span, length, etc..  In the hang gliding context, if the intention was really to say "geometric anhedral" rather than "negative geometric washout", this might mean that we are measuring the downward "droop" in leading edge tubes with the keel tube positioned to be exactly horizontal, rather than with the glider in some other pitch attitude.  This is also often called "airframe anhedral", as it is defined only in relation to the metal airframe, not in relation to the shape of the sail (the fabric part of the wing.)
There is actually one rather tricky nuance at play here.  The geometry of the swept, tapered, "billowed", sail (wing) of a flex-wing hang glider is such that when the wing is loaded, as well as when the trailing edge is simply held up by the "luff lines" with no load on the sail, when we trace the trailing edges from root to tip, we see that they rise up as they approach the mid-span area of each wing, and then they descend again as they approach the wing tips. This is a consequence of sail "billow". In a wing with no taper (i.e. with a constant chord width), and linear leading edges, this kind of trailing-edge geometry would actually mean that the outer part of each wing does in fact have negative washout compared to the mid-span area of each wing. But hang glider wings always have enough taper, in relation to the amount of sail "billow", that the washout actually remains positive, with the twist angle steadily increasing all the way out to the wingtip.  This is a difficult concept to convey in words-- a video walk-around tour of an assembled hang glider would convey the point much more clearly.  Basically what's going on is that in the outboard part of each wing, the trailing edge is descending as we approach the wingtip, but the wing chord is decreasing rapidly enough that this geometry is still compatible with a steady increase in (normal, positive) washout. See also the graphical content on washout due to "conical sail billow" (as opposed to "cylindrical sail billow") in Mark Markowski's vintage (1977) classic "The Hang Glider's Bible"-- this geometry is much is easier to see in the battenless, highly billowed, highly tapered (long wing root chord, short wing span, short or zero wing tip chord) wings of that era than in modern flex-wing hang gliders.  Anyway, it's conceivable that your source is using the phrase "negative washout" to refer to the descending geometry of the trailing edge from the mid-span area of each wing to the wingtip, but if so, this is certainly an incorrect usage of the phrase.  Lay a straight-edge (ruler) from leading edge to trailing edge on the underside of the wing of any assembled flex-wing hang glider, at various spanwise "stations" from root to mid-span area to wingtip, and you'll see that the washout is in fact in the "positive" direction, steadily increasing all the way from root to mid-span area to wingtip, despite the descending line of the trailing edge in the outboard portion of the wing.
Don't entrust your life to poorly-written manuals!  Hang gliding has reached a high "state of the art", and the optimal configurations are well known.  Speak to a manufacturer or dealer for your specific wing.
Footnotes:

It's a function of simple geometry that when we vary the pitch attitude of a swept-wing aircraft, we also vary the amount of apparent dihedral or anhedral that we'll "see" if we view, photograph, or measure the leading edges of the aircraft, as we stand somewhere along the extended centerline.  Hence the several different suggested ways to assess dihedral or anhedral.  Ultimately what matters is whether or not the existing geometry creates a difference in angle-of-attack between the left and right wings in the presence of sideslip, and this may be relatively independent of the overall angle-of-attack of the entire wing system.  It appears there may be merit in assessing the geometry for dihedral or anhedral when the wing system is in the pitch attitude that yields a net lift force of zero when the relative wind is horizontal.  All these considerations are completely separate from the issue of the roll torque directly created by sweep itself, independent of dihedral or anhedral, during a sideslip-- the roll torque created by sweep itself is highly dependent on overall lift coefficient, which is highly dependent on the overall angle-of-attack of the wing system as a whole.

On a calm day, one way to make the airframe and sail of a flex-wing glider adopt a position slightly closer to the actual in-flight configuration, is to simply stand the glider on the kingpost and hold it there with helpers (or ropes).  The empty weight of the airframe and sail will load the lower wires, unload the upper wires, and hold the sail in a shape that somewhat resembles the actual in-flight shape, though with significantly less "billow" than would be seen in actual flight.

Because when the glider is unloaded on the ground, the "luff lines" pull the sail (wing) up into the intended "billowed" shape.  And see comments above re illustrations in Markowski's book showing how "cylindrical sail billow" contributes to washout.

PS - may be able to find some links to photos that illustrate these concepts more clearly- and maybe even a couple scans from Markowski's book-- but that's way on the "back burner" for right now.

@Entaldalpe - answer has been over-edited already, but would have good to emphasize that - unlike anhedral/dihedral- washout is not really much affected by the adjustment (length) of any particular cable, at least under normal flight loads. "Luff line" comments only apply when wing is unloaded on ground. Hence my strong suspicion that your source really was intending to discuss anhedral, not washout. Exactly what does control the washout angle, in a flex-wing hang glider in actual, normal flight, where sprogs and/or luff lines are inactive, is a complex question. Batten tension plays a role.

PS by "mid-span" I mean about halfway out from root to tip-- not at the aircraft centerline.

Washout is a term for twisting a wing.  Washout is measured such that the tip of the wing's leading edge has rotated 'down' as compared to the root.
Washout serves to un-load the wing tip.
Geometric twist is exactly that -- changing the angle of the airfoil along the wing section.
Aerodynamic twist is when you use changes in camber to achieve the same kind of un-loading of the wing.  For example, a root section with substantial camber and a tip section that is symmetrical.
Negative washout would simply go the other way -- increasing the local angle of attack of the wing as you move from root to tip.

"Negative washout would simply go the other way -- increasing the local angle of attack of the wing as you move from root to tip."-- this is a true statement-- but  no hang glider would ever have this feature.

@quietflyer I agree it smells as fishy as can be.  I was just trying to answer the question as concisely as possible.  In particular, the emphasis of geometric twist vs. aerodynamic twist.


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Is a magnetic compass required for VFR (day) flight in powered aircraft?
As the title suggests, Is a magnetic compass required equipment for VFR (day) flight in powered civil aircraft? I have always been under the impression it was.
My question arises from reading CFR section 91.205, article b:

b) Visual-flight-rules (day). For VFR Flight during the day, the
  following instruments
         and equipment are required:
...
     (3) Magnetic direction indicator

This sort of implies I can use anything, including my GPS that indicates magnetic direction. Is my interpretation correct? Has this changed recently?

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

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Should you trust the POH in this regard?
The pilot's operating handbook (POH) says my take off and landing distance are 945ft and 550ft for landing.
Is it wise to increase those factors to 1000ft for safety?
Will I be penalized on the oral if the numbers are not accurate to the POH but adjusted because of safety?
These numbers are baseline numbers not taking into consideration for headwind and tailwind etc…

Yes you'll be penalized because the question is to determine if you know how the derive the numbers from your POH, not how much additional fudge factor you think is necessary.
The examiner should specify the parameters, such as altitude, temperature and winds, and leave it to you to specify the "official" value from the POH's performance chart or table based on the parameters specified, and you'd answer that, according to the POH, for the conditions specified, the numbers are X and Y.
Then the examiner may throw a ringer in, with a follow up question like, "The runway is 950 ft, but has a significant uphill slope, not accounted for in the performance chart. Can you still go?"  That's when you'd say no, you'd want additional margin to account for the slope.
Or the examiner may say, "You're landing on a grass runway that is 575 feet, but the grass is wet and the POH only gives dry runway data or doesn't specify. Can you land safely?".  And you'd explain that wet grass has very poor braking friction (about like compacted snow) and the POH minimum is not enough.
You're not expected to know exactly how much safety margin to add when the additional information is not provided, just show that you know how to derive the performance data from what is provided, and where those sorts of follow up questions are asked, understand how conditions not accounted for in the POH can change the calculation. The examiner is looking to see if your training was reasonably comprehensive, and you can think beyond the basics.

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The Fokker Universal has a crank you have to spin rapidly upon startup. What is it?
The Fokker Universal is one of my favorite aircraft. There is only one complete representation of it in existence at a museum in Canada. A video series I saw on YouTube tells the story of its restoration and some of the history associated with it, and I found it all very interesting. But what confused me the most was the startup procedure.
I know the engine is mainly started with an inertia starter, but I have no clue what the other crank in the cockpit is. If anyone has info, please help.
This first video is part of the video series I mentioned earlier. At 0:15, the crank I’m asking about is shown:

This second video shows the inertia crank on the outside of the aircraft. It shows up at 3:52:

@PcMan -- that does seem likely.  And if so, my answer is not really accurate-- the cranking is not really "engaging" the flywheel clutch, but rather is playing some other role in the starting.  But why the need for the generator?  Is the generator somehow "helping" the magnetos to generate spark?  That's the most likely thing I can think of at this time--

@PcMan -- you should answer then-- you are welcome to borrow any portions of mine you wish; I anticipate I'll be deleting it eventually--

@PcMan (Some aircraft use an "impulse coupling" to give the magneto a faster motion and boost the spark that way during cranking; aircraft that use this method don't typically don't give the spark a boost from any other external source.  And I don't believe that your typical 1950's-technology light plane with electric starter is tying the battery to the spark in any way.  But I am sure you are right about that hand crank in the cockpit.)

Thanks for posting these videos. I looked up the Universal in my copy of the 60-year memorial edition of Anthony Fokker, it mentions that there was a fully restored F.XI in Melbourne, owned by Ansett, in 1979, called Reginald Ansett's first Fokker Universal.

It's being started with an electric inertia starter but the hand crank is not directly related; it's below the magneto switch and is part of the ignition system.
Before magneto impulse couplings that provided a decently strong retarded spark for starting, mags used a separate primary ignition circuit with its own retarded-timing ignition points, for starting only (the normal points operate with way too much advance for starting), which didn't run off the magneto's internal generator and had to be powered externally.
The most common system was called a vibrator, or "Shower of Sparks" (a trade name) ignition system, which used an oscillating current device that would send shots of current through the primary coil to generate a succession of sparks starting at about Top Dead Center of a cylinder when energized from the battery.  When the engine was running, the starting circuit was simply inactive.
The system in the video is similar but runs off an independent Permanent Magnet Generator turned by the crank to make sparks that are suitably retarded.
So what you are seeing is the engine is cranking, the normal magneto circuit isn't really making any significant spark because it's spinning too slowly, but at the right time the little cranks is spun, which will supply sparks, through the separate primary starting circuit, to light off the engine, and as soon as it's running, the normal mag circuit takes over. That's why it cranks, but doesn't fire until the crank is spun.

Yes but the question is what is going on in the cockpit at 0:24 to 0:27 in the first video?  The pilot is rapidly turning a crank.  It appears to be related to engaging the (already spun up) inertial starter, but why would this be accomplished by rapidly turning a crank?

Yeah I didn't read it carefully,  Rewritten as a new answer.

At least some (all?) PT-13 Stearmans with 220HP Continental engines used for primary training in WW2 had such starters. I remember as a boy cranking them up after they came into civilian use after the war. After you had them wound up, you engaged the starter by pulling a T-shaped lever.


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Without wind, would a plane go straight if the pilot let go of the controls?
If you let go of the steering wheel of a (properly maintained) car, it will typically go in a straight or nearly straight line. Would a powered airplane in no-wind conditions do the same thing?

Consider this example, where the pilot not only lets go, but departs the aircraft!  https://en.wikipedia.org/wiki/Cornfield_Bomber

@DJohnM - post cornfield, I worked on that particular jet at Griffiss AFB, 49th FIS. It now resides at the museum at Wright-Pat.

Can you say why planes might not do that?
Are planes not bound by the first law of motion, saying that a body continues in a state of rest or uniform motion in a straight line, unless acted upon by a(n external) force?

@RobbieGoodwin gravity is an external force, and I didn't know if the plane would stay up on its own. Gravity (mostly) isn't an issue for cars in this case because they're supposed to be down on the ground.

It's good that you realised gravity was an external force and then what?
Whether a plane might stay up on its own depends on gravity only if all other things are equal. Otherwise, it depends on the inequalities.
Yes, if you let go of a car's steering wheel it will typically go in a (nearly) straight line.
Yes, planes do the same, the difference being the number of dimensions.
Cars go forward or back, left or right but neither up nor down.
Planes go forward or back, left or right and both up and down.

@RobbieGoodwin exactly; I assumed a plane would keep going forward and not turn left or right if the pilot let go of the controls, but I didn't know whether or not it would stay at a constant altitude, because that's a factor that isn't relevant to cars, and I didn't know how gravity or the lift of the plane would affect it. Sorry if my questions don't make a lot of sense; I don't know much at all about how planes work.

@Someone When you assumed a plane would keep going forward and not turn left or right if the pilot let go of the controls, but you didn't know whether it would stay at a constant altitude, why did you let cars limit your thoughts?
When you didn't know how gravity or "the lift of the plane" would affect it, why would it not have been a good idea to compare the plane's movement in three dimensions to a car's movement in two? Why would "up or down" differ from "forward or back" or "left or right"?
If those queries don't matter, why are you Asking?

@RobbieGoodwin Up/down is fundamentally different from left/right or forward/back (at least where gravity exists), because gravity is creating a downward force external to the object. Without wind, there is no such force in the horizontal plane (no pun intended).

“Without wind, there is no such force in the horizontal plane”.  Please read the second paragraph of my answer and try to let the meaning sink in.  (Hint:  Imagine standing in the gondola of a hot air balloon drifting over the ground at 10 knots.  Would you feel 10 knots of “wind” on your face?)

@MichaelHall so "without wind" should be "without changes in wind speed or direction"?

@Someone Can you say what difference it makes if we're looking at a powered plane or a glider?

https://en.wikipedia.org/wiki/Paper_plane

@RobbieGoodwin a glider will, well, glide down. It certainly can't stay level. Apparently most powered planes do stay level without input, but I didn't know that before asking this question.

@Someone Military gliders are designed to fall safely - though they often don't. Civilian gliders are designed to use the air currents around them, the idea being to rise: ask a glider pilot.
They do that by using what you might call "up-winds", generally know as "thermals."
If you see thermal up-winds as different to left or right or head or tail winds, how does that work?

@RobbieGoodwin so a civilian glider would just fall, not glide, with no wind?

@Someone Absolutely not. Why would you think that off your own bat, let alone when I'd already told you civilian gliders are designed to use the air currents around them, the idea being to rise: ask a glider pilot.
Why do you not accept that they do that by using what you might call "up-winds", generally know as "thermals." If you see thermal up-winds as different to left or right or head or tail winds, how does that work?

We are getting into the weeds here, but effects of wind depends on whether by "go straight" you are asking about maintaining a constant ground track, or a constant heading.  (Or a constant altitude if you are considering vertical "wind" in the form of thermals, updrafts, turbulence, etc.)

@RobbieGoodwin, can you explain what you mean by:  "Military gliders are designed to fall safely - though they often don't."?  P.S.  The OP didn't say they didn't accept that glider pilots could use thermals to gain altitude, only that they weren't aware.  Not too uncommon or surprising if your glider experience is limited to throwing paper airplanes...

@RobbieGoodwin I specified "without wind" in the title, so any scenario involving wind is irrelevant. I was thinking of powered planes, not gliders, although I did not specify that.

@MichaelHall is right; I believe you when you say gliders use thermals to climb; I just didn't know that before you said it.

@MichaelHall Risking exile to Chat, do you see no difference between great and little details of gliding, and Someone's Question?
Broadly, civilian gliders are crafted for the joy of flight; military gliders to be towed to a target then released to fall to ground… often, disastrously.
OP did say (he?) wasn't aware gliders could use thermals to gain altitude and if one is approaching a Question like this from a perspective like that, why should one not be expected to treat up and down in the same way as left and right or for and aft?
Is this about experience, logic or neither?

@RobbieGoodwin, I don't consider chat to be exile, but rather an excellent forum for continuing fruitful conversations!  That said, I'm not totally sure of your point...  I am familiar with historical military gliders, but I don't believe them relevant to the question at hand because I am not aware of any being in use in the last 3/4 of a century or so.

@MichaelHall Good for you, and where did your "3/4 of a century or so" rule originate, please? Why would anyone mind?
I was working on known historical experience. What were you working on?

@RobbieGoodwin even without wind, up/down seems like it would be different because of gravity. (My pronouns are he/him).

Yes, of course. The point is that with or without wind, up/down will be different because of gravity.
If you think back to first principles, how might up/down be different because of gravity?
At the same time, how might left/right or fore/aft be different because of gravity?
Are those principles muddy?

Why are telling us "My pronouns are he/him", please? If you have doubts about your gender or sexuality, are there not many counsellors available?
If you have no doubts, why not use ordinary English?
If you see a third possibility, what is that, please?

@RobbieGoodwin you said "OP did say (he?) wasn't aware gliders"; I thought the parentheses and question marks meant you weren't sure which pronouns to use, and I was confirming that you got them right. I'm not questioning my gender; I was just clarifying something I thought you weren't sure about.

I wouldn't expect fore/aft and left-right to be affected by gravity if the plane is level and balanced.

@RobbieGoodwin, I have no "rule", so I don't know what you mean...  Are there any military gliders currently in use?  Whether there are or not, what is the relevance to this question?

@RobbieGoodwin what did you mean by putting "he" in quotes with a question mark? I never tell people my pronouns normally; this was the first time I ever did it, and I did because I thought you asked.

@MichaelHall You said you don't believe historical military gliders to be relevant to the question… not having being in use in the last 3/4 of a century or so. Not be facetious, if that wasn't as close you could get "ruling them out", the time lapse would not change the way mechanics or physics work.

By the way, I'm sorry I didn't notice this before and I, for one should have been saying "straight and level" throughout.

Gliders in general aren't really relevant to my question; it's about powered airplanes.

Yes, what the OP said.  And not considering something relevant to a question due to length of elapsed time, limited sample size, history of failure, etc. isn't the same thing as a hard and fast "rule" as you described it.  But if there is something specific about military gliders that you believe enhances our understanding about aircraft stability then please write it into an answer.

Broadly, it is a simple question about stability.  (Are airplanes stable?  Y/N)  Some background on different types of stability, or trim, is appropriate in any answer, but I it wasn't a category or class question and I think we can let this go...

P.S.  @Someone, I notice you haven't accepted an answer, do you feel any are lacking?

@MichaelHall thanks for the reminder; I accepted one.

Just to add something to the already given answers.

Hands-off, an airplane is stable but it moves a bit around this stability.
As already explained, if the airplane is stable (inherently or artificially) and trimmed then it is going to fly just dead straight. Anyway even within this boring flight path there exist "stable-instabilities" which are not necessarily triggered by wind/gust and which are going to give a

nearly straight

flight path.
Let's consider a jetliner and let's say that, due to whatever reason (except obviously wind/gust: pilot nudges on the pedals, turbolence over airplane's surface is not perfectly symmetrical among left and right side, ...), it yaws a bit nose-right. The following picture shows this  yaw in an exaggerated manner:

What happen now? Well, as said the airplane is stable so it somehow reacts to this initial yaw trying to return to its straight position... but not that simply. Qualitatively:

Drag and lift generated on the left wing possess now a bigger leverarm in respect to the CG and therefore the left wing tends to go up (bigger lift's leverarm) and back (bigger drag's leverarm) creating a left-up roll and a yaw opposite to the initial yaw.
The left wing rotates/moves against the airflow and therefore its local airspeed increases, while on the right wing the local airspeed decreases. Bigger speed means bigger lift and drag and therefore the left wing tends to go up (bigger lift) and back (bigger drag) creating a left-up roll and a yaw opposite to the initial yaw (just like in 1.)
The vertical stabiliser possesses now an AoA different from zero and produces a lateral force. This lateral force is also acting against the initial yaw but!
It is also located above the CG (seen from the back) and it creates therefore a left-up roll as well.
Due to this roll, the local AoA on the lowering right wing increases while on the rising left wing, it decrease. Lift becomes therefore bigger on the right wing and this bigger lift opposes to the initial roll and levels the airplane off.

That's a lot of events for only a light nudge on the pedals. And it is not yet the end of the story: these restoring yaw and roll don't make only the airplane go back to a straight flight but they actually make the airplane overshoot it's initial straight condition. The airplane continues the yaw and the roll movement on the other side! And then, since it is stable, backward again! This nice picture depicts this "stable instability":

It is called Dutch roll and it is one of the few typical and well known "dynamic modes" of airplanes.
So, hands-off an airplane is stable but it moves a bit around this stability.

If the airplane is properly trimmed, the airmass is smooth, and the aircraft is inherently stable in its design, then yes - many airplanes are capable of flying straight with only very light and occasional inputs.  In fact pilots are trained to trim so that very little correction is needed to maintain straight and level flight.
P.S. "Wind" is simply the relative movement of a mass of air across the surface of the earth.  This concept is lost on the airplane, that knows only its own movement through this airmass.  The airplane responds only to instability of the airmass in the form of turbulence, gusts, or windshear, all of which have the potential to upset its equilibrium.  The conditional statement "without wind" is actually irrelevant to the question.

Read "without wind" as "without gusts" and the term becomes relevant indeed. +1

@Peter, your comment is also relative. In the frame of reference of the airmass, gusts are also irrelevant. Changing the velocity of the wind (the relative velocity of the airmass frame of reference with respect to the ground) will not change the direction of a body moving in the airmass. And even if you fly through a wind shear where the airmass suddenly starts moving in a different direction, the inertia of a body with mass causes it to continue moving in the direction it is already moving in. (One of Newton's laws) It will simply weathervane to align itself with the new relative wind.

Of course, if you mix metaphors (or concepts) by performing mental analysis in more than one frame of reference at the same time (the ground, and the airmass), then it is easy to become confused. Considering the path of a body in a frame of reference other than the one the body is moving in creates the perception that it is changing direction, when in its frame of reference, it is simply changing the magnitude of its velocity, (slowing down) not its direction. Throwing something out of a moving car is a great example. @Michael, your P.S. is spot on.

@CharlesBretana No, gusts are relevant because the airplane does not fly into them all at once. Instead, local angle of attack varies over length and/or span, depending how the gust is shaped and how it is entered and exited. This will upset a trimmed airplane.

I would lump gusts and turbulence together.  Both contribute to unstable air upsetting the aircraft.  That was my point anyway, I think we may be splitting hairs here…

The more abruptly an aircraft transitions from one airmass into another moving at a different velocity relative to the ground (read a gust), the more rapidly the gust will affect the aircraft.  But regardless of how rapidly the transition occurs, it will not (it cannot), change the inertial direction of motion. Aircraft have mass, therefore they have inertia, which means that unless a force is applied to them, they will maintain their inertial direction of motion. Any change in wind will only cause a weathervane effect. It only changes the direction it is pointed in.

... and the only force that might cause such a change in direction would be from the sideways force imparted by the relative wind when the aircraft is in a sideslip. And this force, when adjusting to a gust, is small, (it is dependant on the magnitude of the sideslip angle), and extremely short- lived.

Worth noting that many large aircraft exhibit extremely slow spiral mode instability due to the large vertical stabiliser, which is there to damp Dutch roll (which tends to make people sick!).  So with no hands on the controls, such an aircraft will very slowly enter a tighter and tighter spiral.

I think gusts are very relevant. An aircraft should be stable and return to it's equilibrium after a small disturbance, but if the disturbance is big enough, every aircraft will be unstable.

If "fly straight" refers to track, then wind can make an aircraft not fly straight.

@Acccumulation, I agree.

Yes. Most aircraft are designed to be inherently stable.  That is, if you just let go of the controls, the aircraft will return to whatever it is trimmed to - typically stable and level flight.
Trims are redundant controls (rudder, elevator, ailerons) which are controlled by a "set-and-stay" knob rather than a dynamically movable yoke or pedals.  Turning the elevator trim knob has the same effect as pushing or pulling the yoke, for instance.  So if the pilot wants to hold a rate of climb or descent (often 0), they set the trim knob rather than use muscle to hold the yoke, which would be fatiguing.
If the trims are used conventionally, the pilot doesn't need to hold the yoke at all, and the airplane will keep doing what the trims are set to do, typically straight and level flight.
Generally, aircraft which are not inherently stable will inherently fall off-center, which will cause a vicious cycle and will rapidly exceed flight maximums.  They must be kept at center by constant correction.  The workload involved is generally too intense for pilots, so they must have some sort of automation to do it for them.

Gliders will often show unstable phygoid and spiral dive modes. Still, pilots will fly them without problems because those instabilities evolve slowly.

Large jets tend to exhibit unstable spiral modes too because of their large vertical stabilisers, but again, the modes are so slow that they pose no problems for a pilot.

The real answer is that it would fly fairly straight. This was realised by the Wright brothers who were bicycle makers.
Both for bicycles and aircraft you have a choice between stability and manoeuverability. Some very early aircraft were too stable. Set on a straight course into a hillside they would follow it. The ideal is to be slightly unstable with a long natural period. That way a straight course can be imposed, by pilot or autopilot, with very little exertion, but if a turn is needed it can be done quickly

If the airplane is configured to fly straight and level, then it'll remain that way until the pilot decides to touch the controls. This is especially true if the aircraft has an autopilot system. In a no wind scenario, the airplane will settle in straight and level flight at a particular angle of attack at a given airspeed. As long as the aircraft elevator can be trimmed to maintain this angle of attack, then the pilot will not have to give pressure to the yoke. If the air was not 100% calm, then adjustments may have to be made, but depending on the positive stability of the aircraft, it'll stabilize and regain straight and level flight. So the answer to your question is, generally yes.

There are many airplanes which will eventually tend to slowly roll into a banked turn if the pilot lets go of the controls (and takes his or her feet off the rudder pedals) in smooth air. If the aircraft is truly perfectly trimmed in roll, and the air is free of gusts (not wind-- a steady wind is irrelevant), then the direction of the roll and turn (left vs right) will be random.
An airplane's tendency to stay wings-level with no control inputs is called "roll stability" or "lateral stability".
Design features that tend to enhance roll stability include dihedral, sweep, and a high-mounted wing.  Design features that tend to make an aircraft unstable in roll ("spirally unstable") include a low-mounted wing, a long wingspan (because if the airplane starts to turn at all, the outboard wingtip describes a larger radius and thus experiences more airspeed and generates more lift than the inboard wingtip), and a very large vertical fin (because it minimizes sideslip in turning flight, and sideslip is fundamental to how dihedral, sweep, and high-wing placement all contribute to roll stability.)
Examples of aircraft that will eventually roll into a banked turn in the absence of control inputs from the pilot (or autopilot), even when flying in smooth air and trimmed to be perfectly balanced in roll, include nearly all sailplanes (and motorgliders), and all purpose-designed aerobatic airplanes with a mid-wing configuration with zero dihedral.  All modern hang gliders also exhibit this behavior -- earlier "Rogallo" designs did not, and would stay wings-level when flown for extended periods without the pilot touching the control bar.
In an aircraft that tends to roll into a turn when flown hands-off, as the bank angle increases, the airspeed (and therefore the wing loading, which is essentially lift force per unit weight) tends to increase.1  The flight path tends to descend.  The result is often called a "diving spiral".  In some cases-- but not in all cases-- in the absence of further control inputs from the pilot, the bank angle will tend to reach a maximum value and stop increasing before the aircraft is damaged.  In other cases, in the absence of pilot intervention, the aircraft is likely to be damaged or destroyed either due to overloading, or due to aerodynamic flutter from overspeeding.
I suppose that in theory, if the air were absolutely smooth (free of gusts), and the test were initiated with the wings perfectly level, an airplane that was trimmed to be perfectly balanced in roll would not ever enter a turn, even if it had a configuration such that it was inherently unstable in roll.  But those conditions are never present in reality.  To expect this to happen in the real world, would be like expecting a nail balanced on its point on a hard marble slab to stay vertically upright, because there is no reason for it to fall in any one particular direction.  In the real world, to stay wings-level indefinitely in the absence of corrective inputs from the pilot or autopilot, and not eventually roll into a turn, an airplane must have positive lateral stability (roll stability), and not all airplanes meet this criterion.
There's a more fundamental point at play here though-- up to this point, this answer has addressed an airplane's tendency, or lack thereof, to roll into a banked turn.  But even in an aircraft with strong "lateral stability" or "roll stability", that will roll back to wings-level after any temporary disturbance, there is no tendency for the heading to stay constant over the long run.  Each time the bank angle is slightly disturbed, the heading will change a little bit as the airplane turns slightly.  Expecting the heading to stay perfectly constant over the long run, is like flipping a coin a hundred times and expecting to come up with an exactly 50/50 split between "heads" and "tails".  Just as if you had a car that was correctly aligned so that it had no tendency to turn-- if you drove out on a flat dry lakebed where you could let the car run for miles on end without touching the steering wheel, you wouldn't expect the car to to end up heading in exactly the same direction it started out on, unless perhaps it was following a rutted track left by another vehicle.  In essence, an airplane with positive "roll stability" or "lateral stability" "knows" which way is up and actively returns to wings-level after a disturbance, but it has no way to "know" which direction it was travelling before that disturbance.2
Footnotes:

The angle-of-attack actually tends to decrease somewhat as the bank angle increases and the flight path curves, but not nearly enough to stop the airspeed from increasing.

An anecdote-- I once watched a novice radio-control flyer fly his model airplane-- he was constantly trimming the rudder, and I asked him why.  He said "well, I want it to keep flying into the wind when I take my hands of the sticks, but it won't.  It wants to turn out of the wind."  He thought that he could trim the plane to hold a desired heading, and he was mis-interpreting the plane's random deviation from the desired heading as a specific tendency to turn downwind.  It doesn't work that way-- the best the pilot can do is to trim out any tendency to enter a banked turn, so that it will tend to return to wings-level after any slight disturbance.  He can't trim the plane to hold a specific heading.  By changing the rudder trim whenever the airplane randomly wandered to a slightly different heading, the pilot was actually interfering with, not enhancing, the airplane's inherent "roll stability" or tendency to return to wings-level after a disturbance.  A pilot should trim to remove any consistent rolling or turning tendency, not in response to any random deviation from the desired heading.

If we want to really get carried away, we could consider cases where an airplane with neutral or negative roll stability is trimmed to be perfectly balanced in roll, and air is perfectly smooth, and wings are perfectly level, to such a perfect extent that there is absolutely no initial roll/turning tendency (like a pin perfectly balanced on its point), but after flying a sufficient amount of time in an appropriate direction with zero pitch, yaw, or roll rotation as seen from a valid inertial reference frame, the earth's rotation has caused the airplane to become banked relative to the horizon!


text:
When was 91.155(b)(2) (night VFR cloud clearance and visibility in Class G airspace in traffic pattern) implemented?
We have all memorized the helpful triangle, but now this paragraph:

(2) Airplane, powered parachute, or weight-shift-control aircraft.  If the visibility is less than 3 statute miles but not less than 1 statute mile during night hours and you are operating in an airport traffic pattern within 1⁄2 mile of the runway, you may operate an airplane, powered parachute, or weight-shift-control aircraft clear of clouds.

... is mucking up the works. When was flying below 1200 AGL in a traffic pattern no farther away than 1/2 a NM in Class G at night and only needing to see 1 NM away & staying clear of clouds deemed a good (legal) idea?

The basic content (without references to powered parachute or weight-shift-control aircraft) was already present in the 1991 Final Rule "Airspace Reclassification", as can be seen by clicking on the first reference link at the bottom of the page for the link that you include in your question.  See page 65660 on this PDF from the Federal Register.
But the rule is older than that--
I have a copy of the 1978 FARs, and at that time, FAR 91.105 governed basic VFR weather minimums.  Only 1 mile visibility was required for VFR airplane operations in "uncontrolled airspace"-- with no distinctions at all between daytime and night requirements.
The FAA increased the basic VFR visibility requirements for nighttime airplane operations in "uncontrolled airspace" in a Final Rule entitled "Night-Visual Flight Rules Visibility and Distance From Clouds Minimums", effective November 13, 1989, issued September 25 1989, and published on pp. 40234-40327 of the September 29 1989 Federal Register (PDF downloadable here).  FAR 91.105(a), on p. 40326, requires 3 miles visibility for airplane night VFR operations in uncontrolled airspace.    FAR 91.105 (b)(2), on pp. 40326-40327, contains the exception allowing airplanes operating under VFR in uncontrolled airspace in a traffic pattern within 1/2 mile of the runway to operate with visibility of less than 3 miles, so long as the visibility is greater than 1 mile.
In the discussion section on p. 40325 of this document, you can read about the AOPA's support for allowing operations at night in uncontrolled airspace "under the provisions of the current rule", e.g. with visibility as low as one mile in uncontrolled airspace, as long as those operations are conducted in a closed traffic pattern, to "allow pilots to maintain currency and proficiency in night takeoffs and landings per § 61.57(d)."  FAR 91.105(b)(2) was essentially an accommodation of this viewpoint, and today's FAR 91.115(b)(2) is the direct descendent of that.

Folks' typical idea of the appropriate size of a traffic pattern for closed-circuit night VFR operations has perhaps increased a bit since 1989?


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How do I appropriately choose the power settings on a C172 for cruise?
Today during the ground portion with my instructor he asked me "why did you choose to cruise at 2300 rpm flying at 1990 ft pressure altitude?" I didn't have an answer for him, and I couldn't find anything in the POH or online that could answer that question.
All I was able to find and surmise is "cruising should be done between 55% and 75% power, it is up to the pilot's discretion to choose the power setting in this range for the planned route"
So now I'm stuck with this in my head:" As long as my power setting allows me enough range and endurance to reach my destination with reserves in mind and that my cruising speed is not ridiculously slow any power setting between 55% and 75% should do"
However, previously another instructor told me for a C172 a power setting between 62% and 65% is the best balance between performance and engine care.
what is the right way of going about this?

Look at gallons per hour vs speed and altitude in the POH.  Then decide if range, terrain,  time, weather, or local rules affect the planned route.

I think there may be a typo or grammatical error here.  Was this really the word you wanted to use?  https://en.wikipedia.org/wiki/Digression  .  Feel free to roll back my edit if your word choice correctly reflects your original source-- but consider providing a citation--

While the choice of altitude and power is indeed at your discretion, you should be able to explain why you made the choice you did.
Best Speed
Assuming range isn’t a problem, you
typically want to get to your destination as quickly as possible, especially when you’re paying a wet rental rate.
Look in the cruise performance table to find the best TAS for each potential altitude, and then factor in the head/tail wind for each altitude according to the winds aloft forecast.  The altitude and power setting with the best ground speed wins!
(Technically, you also need to factor in the extra time it’ll take to climb to a higher altitude, but it’ll only be a few minutes, so you can skip that for now.)
Best Range
Flying faster burns fuel faster. If you’re going a really long distance, it may burn so much fuel that you won’t make it!  Maybe you’re better off flying slower to get more range.  Or maybe you’re renting dry and just want to save fuel. Or you’re the person paying the maintenance bill.
Do a similar calculation to the above, but instead of maximizing ground speed, you’re minimizing fuel per mile.  Or at least getting it low enough that you can make it to your destination (plus reserves).
Note that even if you can stretch the plane to fly five-plus hours per leg, do consider whether you can handle being cooped up for that long without a break. You may prefer to fly faster with fuel stops that will double as an opportunity to get out and stretch—and use a toilet.

Or a smart compromise between the two.

Bottom line: if it's a wet rental and you just want to get from A to B on as little hobbs time as possible, wind it up to max cruise. If it's your airplane and you're paying for the gas, fly at best MPG if you're not in a hurry.  I cruise my own plane at about 50% most of the time.

Re "Look in the cruise performance table to find the best TAS for each potential altitude, and then factor in the head/tail wind for each altitude according to the winds aloft forecast. The altitude and power setting with the best ground speed wins!" -- so this would always be redline RPM, or full throttle, whichever comes first, right?  If not, can you expand your answer to say why not?

@JohnK -- if you are really not in a hurry, why not fly at best gallons per hour?  After all, the real point of most of the flying that many of us do, is simply to spend time in the air, right?

@quietflyer Lower altitudes usually aren’t WOT at cruise power.

@quietflyer If you’re paying a wet rate (typical for rentals and clubs), you want to fly as fast as possible.  Why spend more to save someone else money?

@StephenS --  re "Lower altitudes usually aren’t WOT at cruise power." -- but what defines "cruise" power, and why is it so defined?  If you are really going for "best speed", why is someone's (arbitrary?) definition of "cruise power" of any interest?  Is it bad for the engine to run at higher than "cruise power" for an extended time?  If so, the answer would benefit from including this info.  In the absence of specific well-defined engine-health considerations, if your priority is saving money on a wet rental over set distance, I fail to see why WOT or redline RPM would not be your target.

Re " If you’re paying a wet rate (typical for rentals and clubs), you want to fly as fast as possible."-- that wasn't true for me, when I was paying a wet rate.  Obviously, using airplane as economical, fast transportation from point A to point B is a real thing, but it's not the actual reason behind most of the air time incurred by private pilots flying light airplanes.  Granted, I could have spent the same and covered more ground, I guess, by flying WOT or redline, but-- the scenery was about the same over the shorter distance versus the longer.

@quietflyer The POH defines cruise power and any limitations, such as not exceeding 75% power for extended periods.  But within those limits, I want to pay as few dollars as possible to accomplish my mission, and a wet rate means a higher ground speed is nearly always cheaper.

@quietflyer I could fly the PL-2 around at 35% and still get 100 mph out of it and only burn about 3 1/2 to 4 us gal/hr, but at some point the high pitch attitude becomes a nuisance, and some engines don't like certain RPM ranges and won't run that smoothly.  My O-290 seems happiest around 20" mp and 2000-ish rpm, which is around 50+% with a burn of about 41/2 us gal/hr.

@StephenS - does pilot's manual of Ces 172 really specifically recommend not exceeding 75% power and/or "cruise power" for extended periods of time?  If so, I overlooked that when I was flying one frequently-- (or else I've just forgotten)--

Thank you. This is the answer I've been looking for. The route i had planned was well within the range of the 172. Since both extreme side of the power setting provide enough range to get to my destination, my instructor said 65% power, which is the average between 55% and 75% would be the best compromise between range and speed.

@YingTaurus If you have the range to get there at 75% power, why pay more to get there slower? That compromise benefits whoever you’re renting from (maybe the CFI’s employer?) rather than you.


text:
How the AOA, Pitch and FPA are changing during a flight maneuver
For a steady state flight, there is a equation that FPA + AOA = Pitch, but how these three variables are correlated during a flight maneuver like currently I am in a leveling flight and then I want to change the flight path angle to 5 degrees, can I assume that AOA will be unchanged during the maneuver, so I could simply pitch up 5 degrees more to get a 5 degree FPA as delta FPA will be same as delta pitch（ assuming the speed is well maintained thru the thrust）?

Your formula isn’t correct.  Does this help?  https://aviation.stackexchange.com/questions/35396/why-the-pitch-rate-q-is-zero-while-angle-of-attack-is-changing-and-vice-versa?rq=1

@quietflyer why constant airspeed indicates constant AOA?

@quietflyer, Don't you think that angle of incidence is just as important to consider as AOA in this formula?  And why would you assume AOA to be relative to the fuselage when the very definition of AOA references the chord?

@MichaelHall -- I wasn't saying that incidence isn't important, just that it's not really germane to the question, because it's fixed.  If incidence is the reason you said the formula was wrong, then technically yes you are correct, I hadn't thought of incidence right away in my earlier comments.  As per AoA relative to fuselage vs wing-- yes normally we define it relative to the wing chord, but see https://aviation.stackexchange.com/q/76683/34686 especially answer https://aviation.stackexchange.com/a/76709/34686

Well I still would argue that the question really should be narrowed, but obviously I've decided that it is worth answering anyway-- here goes--

@VvV, from the way the question is phrased I have to assume you are not a pilot, but would you please explain what you mean by asking how the “variables are correlated”?  It might help someone tailor an answer appropriate to your specific need if we understood your goal, as well as the root of your confusion.  (I would phrase an answer differently between a new student pilot vs someone writing software code for a simulator)

Because the base question is:  "can I assume that AOA will be unchanged?"  Which if you paraphrase becomes "if I pull back on the yoke will the AOA change?" The obvious answer is yes, (to the second question, no to the first...) but the fact that you might presume that AOA doesn't change with pitch inputs indicates a deeper misunderstanding that needs to be rooted out in order to best complete your quest for knowledge in this specific area. Can you help us help you?

@MichaelHall I am a pilot and I am also learning to automatic the aircraft controls, I ask this question as I want to have a controller to control the vertical path simply closed loop with FPA/pitch without knowing the actual AOA.

Thanks.  Knowing you are designing an autopilot for your airplane helps.  I would suggest that you consider pitch rate.  Because for a slow and smooth 5 degree pitch up, (especially compensating with throttle) AOA changes will be negligible.  AOA becomes a bit more crucial in turns and at approach speeds.  And especially at higher pitch rates.

The required lift force is about the same in a shallow climb as in level flight.  (Strictly speaking, a steady-state climb always requires less lift than level flight-- for more see Does lift equal weight in a climb? -- but the difference is very small for climb angles up to about 30 degrees.  At a 5-degree climb angle, lift is .996 * the aircraft weight.)

The amount of lift a wing generates at any given moment depends on the wing's angle-of-attack, and on airspeed.  Lift is roughly linearly proportional to angle-of-attack through much of the flight envelope, and also is proportional to airspeed squared.

Therefore the airspeed indicator can be thought of as an angle-of-attack indicator, at least during non-accelerated maneuvers, i.e. during 1-G flight, excluding very steep climbs and dives.1  If the airspeed has decreased and the aircraft has remained in 1-G flight, the angle-of-attack must have increased.

Also, to a first approximation the control stick or yoke is really an angle-of-attack controller, and the fore-and-aft position of the control stick or yoke serves as an angle-of-attack indicator.  For much more on this, see Wolfgang Langewiesche's famous book "Stick and Rudder".  Generally speaking, if the pilot moves the control stick or yoke aft, the angle-of-attack of the wings is increased.  Of course, thrust changes can produce changes in pitch trim that complicate this relationship-- in some aircraft the angle-of-attack will tend to increase, or decrease, when thrust is increased.  (Also, increases in G-loading make an aircraft tend to trim to a somewhat lower angle-of-attack for any given position of the stick or yoke, due to aerodynamic effects generated by the curving flight path.  This effect is most pronounced in relatively slow-flying aircraft, like sailplanes.)2,3

So generally speaking, unless you are making a specific effort to ensure that the airspeed stays constant (e.g. by adding power as needed), you can't assume that when you pitch the aircraft up 5 degrees, the angle-of-attack of the wings will stay constant, and that's why you can't assume that you'll get a 5-degree change in the angle of the flight path.  On the other hand, if you are keeping the airspeed constant-- which obviously would require an increase in thrust and power as the climb is entered-- then this would be a fair assumption.

Footnotes--

Or more broadly, we can say that the airspeed indicator can be thought of as an angle-of-attack indicator in flight at any given constant non-zero "measured G-loading", where by "measured G-loading" we actually mean only the G-loading component that acts in the "upward" direction in the aircraft's own reference frame. Essentially, what we read on a G-meter mounted on the instrument panel.  Note that this "measured G-loading" would be zero in a steady-state vertical climb or a steady-state vertical dive, as well as in a parabolic "push-over" maneuver.

This tendency to trim to a lower angle-of-attack when the flight path is curving in the "upward" direction in the aircraft's reference frame, can also be described in terms of "pitch damping", i.e. an aerodynamic "resistance" to pitch rotation.

Another factor that complicates the relationship between the fore-and-aft position of the control stick or yoke, and the resulting angle-of-attack of the wing, is pitch rotational inertia.  To initiate a pitch rotation, the tail must generate a net pitch torque, above and beyond whatever pitch torque is needed to balance the wing's pitching moment.  But we don't need to worry about this if we're just comparing steady-state cases.

It is surprisingly hard to find an on-line source giving a graph of airspeed versus angle-of-attack on line.  Such a graph would improve this answer, as it would give a feel for how much the angle-of-attack would change, for various changes in airspeed.

Re "It is surprisingly hard to find an on-line source giving a graph of airspeed versus angle-of-attack on line. Such a graph would improve this answer, as it would give a feel for how much the angle-of-attack would change, for various changes in airspeed. "-- here's one-- https://www.av8n.com/how/htm/aoa.html#sec-power-curve-more -- this answer would benefit from the incorporation of this link-- but not today--

FPA + AoA = Pitch

What is FPA?  FPA is flight path angle.  This is (in modern airliners) a way of autopiloting a descent based on angle rather than rate for a given heading.
So, pitch to the horizon is FPA + AoA, or FPA (where you're going) is Pitch - AoA.  This part is correct.

assuming the speed is well maintained through the thrust

Thrust is the key.  If your plane is staticly stable, adding or subtracting thrust will change pitch.  Any changes in AoA will be transient.
If you are more staticly nuetral (depending on your CG location), pitch change is "helped" with a bit of yoke too (or you'll simply speed up when more thrust is applied).
After completing the manuver, for the same speed, AoA (trim) will be the same$^1$, FPA and Pitch will be different.

can I assume AoA will remain unchanged during the manuever?

Actually, no.  During the manuver pitching motion and velocity changes will affect the angle of attack and lift output, which lead to deviation in the flight path.  The best example of this is a failed loop.  As the pitch goes to towards vertical, airspeed cannot be "well maintained" with thrust.  The nose drops below the intended line of flight until enough speed is recovered to resume linear flight.  If the plane stalls, the deviation is even more dramatic.
Conversely, a gentle increase in pitch (with a careful eye on airspeed) will lead to much gentler (if any) pitching oscillations.  The plane will simply nose up and "climb" above its previous flight path.  If done in a leisurely airliner manner (don't spill the drinks), AoA will not change much.

pitch controls speed, power controls altitude.

Or maybe, trim controls speed, power controls pitch.  But if you are staticly nuetral, these functions are not strictly separated, and the aircraft will not seek out a given speed unless (you guessed it) it is programmed to.
$^1$  for a 5 degree climb, Lift requirement from wing (AoA) will be a hair less because the thrust is also doing some of the lifting against gravity.

I dunno Richard, I don't really see pitch stability dynamics as being germane to the question.  Likewise the dynamics of a failed loop-- Also, an aircraft can be statically stable in pitch but still have a significant trim change, in terms of angle-of-attack, as power is varied.  Really all it boils down to, is whether the airspeed is being held constant, or not, regardless of whether the aircraft naturally does that on its own, or the pilot makes it happen--   just my 2c!

AOA is useful information to have, but a great many aircraft do not have AOA indicators.  It becomes more important as you explore the edges of an airplane's performance envelope, but for the purpose you describe it isn't necessary to know what your AOA is.

...in a leveling flight and then I want to change the flight path
angle to 5 degrees, can I assume that AOA will be unchanged...

No, if you change pitch the AOA will change.

...so I could simply pitch up 5 degrees more to get a 5 degree FPA as
delta FPA will be same as delta pitch（ assuming the speed is well
maintained thru the thrust）?

Yes, depending of course on airspeed and pitch rate.
For the example you describe, if you are at cruise airspeed and want to begin a gentle pitch rate to a 5 degree climb, the change in AOA is negligible and virtually undetectable.  You should be confident in programming your autopilot to perform gentle and gradual cruise maneuvers such as this without worrying that you lack AOA input.
In fact, I was trained to climb in either pitch hold or IAS hold, and descend in VSI hold.  While I’m sure there may be airliner autopilots with AOA hold capability, I don’t see that feature, (or even AOA inputs) being necessary on a home built or experimental GA aircraft.

Seems a relevant question here, would be "is the autopilot using the aircraft's pitch attitude, or the aircraft's actual achieved climb angle (relative to ground?  relative to airmass? measured via GPS?  measured via vsi and airspeed?) to govern the aircraft's performance?"  If any of the latter, then the point doesn't really seem germane to the actual question.  And if the former, it might be worth making that more explicit in the actual answer.

@quietflyer, as usual I am trying to simplify while you are overcomplicating things.  I distilled the question and OP's reply to my query to be a concern about using autopilot to gently climb without an AOA input.  I fail to see the relevance of adding GPS, airmass, and all the other things you mention.  But feel free to edit your own answer and let's see who the OP chooses.

The addition of "In fact, I was trained to climb in either pitch hold or IAS hold, and descend in VSI hold. " added significantly to the answer in my opinion.  Some readers may not even know that pitch hold "is a thing" and thus may not have understood what autopilots have to do with the question but it is more clear now.


text:
Aircraft load factor and body normal acceleration
A load factor is defined as "the ratio of the lift of an aircraft to its weight", i.e., N = Lift/Weight.
Normal acceleration (nz) is defined as the component of the linear acceleration of an aircraft along the body Z axis.
So my questions is: does the load factor N is the same as the normal acceleration nz? I thought normal acceleration computation should be more complicated to include all the forces'(weight, lift, drag, thrust)components acting on the body z axis.

What was the actual question here? Are you asking if your understanding in the final paragraph is correct?

It would seem "the ball" gives an accurate picture opposite of the direction of net aerodynamic forces + the contribution of gravity This is why additional yaw must be added to center it on the z axis in a turn.  So now you have gravity + lateral acceleration + yaw to consider.

@RobertDiGiovanni -- no, the ball just shows the (mirror image of the) net aerodynamic forces (including thrust if applicable), or at least the component thereof that acts in the "plane" of the curved tube.  Ball deflection = aerodynamic actual sideforce.  But, best discussed in comments under one of several existing questions on the slip-skid ball, or in a new chat room.

No, the ball shows "centripetal" and gravity forces as per this
reference.  Always happy to help refresh knowledge of even experienced pilots.

@RobertDiGiovanni -- both statements are true.  If you only specify the "centripetal" force, then you must also include gravity into the calculation to see where the ball will end up.  If you only specify the net aerodynamic force vector, specifically it's direction relative to the aircraft (don't even need to know the magnitude), then that's all the information you need to know where the ball will be.  Don't even need to know bank angle, or whether or not aircraft is doing a wingover or a partial loop.  Don't need to know orientation of gravity vector (vertical) relative to aircraft.

@RobertDiGiovanni -- slip-skid ball now addressed in footnote 4 of my answer.

In math and physics, "normal" means "perpendicular to".  I wonder why "normal acceleration" has been defined to only be the acceleration component acting along the body Z axis.  Wouldn't  an acceleration component acting along the body Y axis also be "normal" to (perpendicular to) the "main" axis of the body, i.e. the longitudinal axis of the body, aka the body X axis?  This comment started out as a suggestion for improving the question, but a little googling around suggested that the terminology used here is in fact standard accepted terminology.

I think the real answer is given here -- https://aviation.stackexchange.com/questions/95313/aircraft-load-factor-and-body-normal-acceleration

EDIT: I modified my answer in order for it to match the modifications in the question.

Unfortunately load factor is almost always explained making the airplane doing some cool manoeuvre à la Top Gun: from there to picturing a sweaty Tom Cruise trying to avoid greyout, is just a short step. And to complicate the matter even more, gravity, Gs, apparent weight and other fancy terms are also drawn into play...
But load factor is just what you have written: lift to weight ratio. As simple as that.
The picture in your question (picture which is now gone and that I repost here under) is a very good example to answer your question:

In this example of a steady turn, the load factor $N$, with $\phi=30°$, is:
$N=\frac{L}{W}=\frac{W/\cos\phi}{W}=1.155$
and that simply means that the wing is producing some 15.5% more lift than the airplane's weight (note: weight, not mass $\Leftrightarrow$ Newton, not kg).

does the load factor N is the same as the normal acceleration nz?

This should be better formulated: load factor is a pure number while acceleration is length/time², so they cannot be really compared. "Normal" is also ambiguous. As a consequence I suppose here that "normal" aka perpendicular is the direction going from belly to canopy and that the question is if load factor and "normal lift divided by weight" are the same.
Well, the answer is easy if you picture the airplane from the side at high AoA (image from Wikipedia modified by me):

By definition, lift is perpendicular to the airflow and in this case its direction is not coincident with the airplane's perpendicular direction z. So the general answer is simply no.

I thought normal acceleration computation should be more complicated to include all the forces' (weight, lift, drag, thrust) components acting on the body z axis.

Theoretically yes but practically it depends on how mutually big their z components are. In the first picture  for the steady turn, drag and thrust are basically aligned with the x-axis and do not contribute on the z-y directions and therefore there's no need to consider them in calculation on z or y.

Bonus material

Where does the 15.5% higher load factor of the first calculation go? It goes in the force $F_r$ which makes the airplane stay in a curved path.

Why is only lift considered in the calculation of N? Lift is by far the biggest aerodynamic force produced by the wing and therefore the highest source of stress by a structural point of view.

Why the ratio lift to weight is considered and not the actual lift? Because aerospace engineers like numbers, i.e. quantities without   dimension: Reynolds number, Mach number,  lift coefficient, drag coefficient, ... Dimensionless quantities cannot be mistaken (lb vs. lbf vs. ft$\cdot$lbf for example) and they make comparisons easier.

In the aerospace world there are at least 3 different reference systems which can be used to do calculations: the one attached with the earth (first picture), the one attached with the body (second picture) and the one attached to the airflow. The one attached to the body can also be further expanded according to where it starts and how it is oriented: it can for example originate in the CG and have the z-axis pointing downward; or it can originate somewhere in front and under the nose of the airplane and have the z-axis pointing upward.

Normal acceleration (nz) is defined as the component of the linear acceleration of an aircraft along the body Z axis.

To a physicist1, the term "linear acceleration" in the above definition would mean the net linear acceleration.
The net linear acceleration is the acceleration due to vector sum of all the forces acting on the aircraft, including gravity.
In the context of coordinated flight at a constant airspeed, with a horizontal flight path, Thrust and Drag are equal and opposite, and the net linear acceleration is the acceleration due to the vector sum of Lift and Weight.
In this case, if we draw the closed vector triangle of Weight, Lift, and resulting net force, we obtain the following:
Lift = Weight / cosine (bank angle)
Net force = Lift * sine (bank angle) = Weight * tan (bank angle)
If we make the simplifying assumption that the wing incidence is such that the fuselage is exactly horizontal, so that the Lift vector is entirely aligned with the aircraft's body Z axis, then to obtain the component of this net force that acts parallel to the aircraft's body Z axis, we "project" the net force onto the body Z axis by multiplying by sine (bank angle).  (Similarly, to compute the net "lateral" force-- the net force component acting parallel to the aircraft's body Y axis, we would multiply the net force by cosine (bank angle).)
To compute accelerations rather than forces, we simply substitute "1G" for "Weight".
This yields the following:
Lift = Weight / cosine (bank angle)
Net force - Lift * sine (bank angle) = Weight * tan (bank angle)
Net acceleration = 1G * tangent (bank angle).  (Note that the net acceleration is purely horizontal.)
az = component of net acceleration acting in body Z axis = 1G * tan (bank angle) * sine (bank angle)
ay = component of net acceleration acting in body Y axis = 1G * tan (bank angle) * cosine (bank angle) = 1G * sine (bank angle)
This yields the following results:
0 degrees bank:
Lift = Weight, net acceleration = 0 G, az = 0 G, ay = 0 G
30 degrees bank:
Lift = 1.15 Weight, net acceleration = 0.577 G horizontal, az = 0.289 G, ay = 0.500 G
45 degrees bank:
Lift = 1.414 Weight, net acceleration = 1 G horizontal, az = 0.707 G, ay = 0.707 G
60 degrees bank:
Lift = 2.000 Weight, net acceleration = 1.732 G horizontal, az = 1.500 G, ay = 0.866 G
It is difficult to see why az and ay figures computed in this manner would be of much interest to anyone.
If we wanted to actually predict the aircraft's trajectory, we would work with the net force, not the body-Z and body-Y components of the net force.
And note that when we compute the az figure as above-- by simply taking the body-Z component of the net force acting on the aircraft-- then the resulting figure does not correlate to what we read on a G-meter (which is simply a 1-axis accelerometer, with the sensing axis aligned with the aircraft body Z axis.)  (And likewise, the ay or "lateral acceleration" figure as computed above bears no relation to the deflection of a slip-skid ball-- assuming the fuselage is pointing directly into the relative wind, the ball would be centered in all of these turns, even though the ay figure is not zero.)
Note also that the acceleration component due to gravity creates no stresses or strains on the aircraft structure, or on the contents of the aircraft, because gravity "acts from within", creating an equal acceleration on every molecule of the aircraft and occupants.
The logical conclusion is that in the aviation / aerospace context, az as described above is usually not what we mean by the "normal acceleration" nz.  In the aviation / aerospace context, the "normal acceleration" nz usually refers to the body-Z axis component of the "felt" acceleration, not the body-Z axis of the net acceleration.
The "felt" acceleration is the acceleration resulting from the net vector sum of the non-gravitational forces acting on the aircraft, not the net sum of all the forces acting on the aircraft.
At rest on the ground, the acceleration "felt" by an aircraft is the acceleration component due to the force of the earth pushing up against the landing gear, with a force of 1 G per unit mass.
In flight, the acceleration "felt" by an aircraft is the acceleration component due to the vector sum of all the aerodynamic and thrust forces, divided by mass.
The "felt" acceleration, not the net acceleration, is what of interest if we are looking at the loads transmitted through the aircraft structure and the resulting stresses and strains.
The body-Z component of the "felt" acceleration, not the net acceleration, is what is registered by the G-meter.
(And the ratio between the body-Y and body-Z components of the "felt" acceleration, not the net acceleration, determines the displacement of the slip-skid ball.  The slip-skid ball is centered whenever the body-Y component of the "felt" acceleration is zero, which is also when aerodynamic sideforce is zero.)
In the context of banked, turning flight at a constant airspeed, with a horizontal flight path, with the wing incidence is such that the fuselage is exactly horizontal, with zero sideslip, all the "felt" acceleration is parallel to the aircraft's body Z axis.  In this case, nz defined as the body-Z axis component of the "felt" acceleration is exactly equal to 1G * (Lift / Weight).
(Naturally, if the flight path is horizontal but the fuselage is not exactly horizontal, then the Lift vector will no longer be aligned with the aircraft body Z axis, and nz as defined above will be less than (1G * Lift / Weight).  As illustrated in this answer.)
References from outside sources:
Here is an example of a technical engineering paper where "normal acceleration" in straight-and-level flight is treated as 1 G rather than 0 -- "A Method for Correcting The Error in Indicated Normal Acceleration Due to G-Sensor Location" by Mi Yi and Chen Mingtai of the Shanghai Aircraft Design and Research Institute-- (link to download PDF)
Note: the designations "az" and "ay" were introduced for convenience for this answer; there is no intent to represent that they are standard terminology.
Footnotes:

A physicist would also object to the use of the word "normal" to mean "body Z axis component of".  "Normal" means "perpendicular to the trajectory".  (See for example this link.)  What force component should properly be considered the "normal force" in a wings-level skidding turn, performed essentially with the rudder alone (plus opposite aileron as needed to keep the wings level)? Or in a control-line model airplane, racing around in a circle with the nose pointing directly into the airflow and relative wind, and with the wings completely level?

Thanks, so in a coordinated turn, assuming the body axis align with the wind axis(or the alpha is small), lift could be considered as the dominate external force acting on the body Z axis, so we could approximate the sensed nz as Lift/mass*g. I think approximating sensed nz by lift/weight should be applied to other cases like aircraft is not in a levelling turn but in a climb or descent turning, not sure if it is still accurate if the turn is not coordinated when there is a side force acting on the body y axis.

@VvV -- I would say, given the constraints in first part of your comment, sensed nz  is exactly 1G * Lift /Weight.  In a climb or dive, however, the lift vector is less than in level flight.  See for example https://aviation.stackexchange.com/questions/40921/does-lift-equal-weight-in-a-climb .

@VvV -- when the turn is not coordinated, this definitely affects nz. For example in a "slip" there is a sideforce that partially opposes gravity and supports some of the aircraft weight, unloading the wing, compared to a coordinated turn at the same bank angle. Extreme case: sustained, linear knife-edge flight (90-degree bank angle). Lift from wings is zero, nz is zero. Conversely in a "skid", nz is decreased, and the wing is unloaded, compared to a banked, coordinated maneuver that would yield the same turn rate. So -- yeah, it's a little complicated!

@VvV -- note that in slip or skid, Lift vector and nz are both affected in same way, so relationship between them stays same--


text:
Is "excess thrust" possible in a steady state vector diagram?
Being recently very much sold on the lift-weight-thrust closed triangle (with weight forming the hypotenuse) model for climbing flight, I proceeded to put the logic through its paces (validation) and got stuck on one fairly basic concept:  excess thrust may be only transient.
Increasing throttle increases speed.  We can initiate a climb with more speed, thrust working against increased drag, and to hold the climb against the tilted lift vector by increasing the lift coefficient, by way of increasing AOA.
Note the with a GA aircraft (25% max thrust to lift), increasing AOA excessively leads to loss of airspeed.  In fact, this is how to stall an aircraft.
Pitch can never be transitioned into hover before stalling.
Can anyone confirm that, beyond applying throttle before the takeoff roll, "excess thrust" exists at all?

"Excess thrust" means "more thrust than required for level flight", not "more thrust than required for unaccelerated flight". I think you may just be stuck on this definition?

@Sanchises that would be excess thrust available. Not meaning to be a trouble maker, but that "lift-drag-weight" vector diagram is in question (for light GA aircraft).

I'll post an answer later. The free body diagram cannot be in question for it would spell the end of Newtonian physics :)

Yes, excess thrust always exists when the airplane is accelerating. But that is not strictly steady state. However, it is perfectly possible to extend the vector triangle by an acceleration term which balances the extra thrust. Same with climb: This is not strictly steady state but can be modelled quasi-stationary because parameters change only slowly. In the end, even unaccelerated horizontal flight is not strictly steady state since fuel is burned off, so the aircraft mass declines over time.

@Peter Kampf extra thrust leads to extra drag, either by increased velocity, or by increased AOA, (or by increased camber).  What I'm getting at, for this level of weight to thrust ratio, is that (not even taking the sin (climb angle) and applying it to the vertical weight vector), is that (in an aerogravimetric enviornment), more thrust equilibrates with more drag.  Its the increased energy of the system (Power) that allows us to go faster or climb.

@Sanchises please, no, not the end of Newtonian physics, it just may not apply to a climbing GA aircraft.  There, static stabilty may remain in effect even when climbing or descending, allowing for larger Lift vectors, not smaller.  A Piper Cub cannot take part of its thrust and apply it for lift.  It needs to use the mechanical advantage of the wing.  However, note that the lift-thrust-weight diagram does work (in its simplest form) for a rocket, as the rocket has sufficient thrust to equal weight and drag.

@RobertDiGiovanni: In the first moment speed is still slow and all drag increase is from the stronger slipstream. In a climb thrust must be higher because weight has a backward component (relative to the thrust line) and some of the weight is carried by engine thrust. The same reason why lift becomes smaller in a climb: The engine takes over a bit.

@Peter Kampf thanks so much for your expertise and comments.   They are always a delight to read and ponder.  All of this is plausible, yet I dissent (for now) as lift renders the plane "weightless". No vector to shift fore and aft.  As the wing essentially provides a mechanical advantage (not shown in the vector diagrams), it may be better to use added thrust for added drag (from increased AOA or speed).

It seems this might be a duplicate of -- https://aviation.stackexchange.com/questions/40921/does-lift-equal-weight-in-a-climb/56476#56476 -- at least it may be answered by the reading the correct answers to that question

Re "it just may not apply to a climbing GA aircraft. There, static stabilty may remain in effect even when climbing or descending, allowing for larger Lift vectors"-- you are mistaken in the idea that static stability is somehow opposed to the closed right triangle of (T-D), W, and L, with W as the hypotenuse and L and (T-D) both non-vertical.  Note extensive reference to constant angle-of-attack in all my answers relating to balance of forces in climbing flight.  That's what static stability "tries" to do-- to maintain a constant angle-of-attack.

Note the reference to having to move the stick or yoke forward if we want to see an increase in thrust result in an increase in airspeed and no change in pitch attitude, in the third paragraph from the end in this answer https://aviation.stackexchange.com/a/82047/34686.  See also third paragraph of present answer https://aviation.stackexchange.com/a/82237/34686 for what happens when we increase thrust without changing angle-of-attack.

Is it possible that good answers to the following proposed new ASE question would help satisfy your unease-- "Give a play-by-play description of exactly what happens to the magnitude and direction of the Lift, Drag, and Thrust vectors, and the airspeed vector, and the climb rate, when we are flying level and then increase Thrust, without moving the elevator.  Assume strong static stability keeps the angle-of-attack constant.  Assume zero down-thrust or up-thrust, relative to the direction of the flight path."

Not sure it that would be a duplicate of any already asked question or not.  Many of my answers touch on this but not sure any exactly go through in full detail, including the forces during the transition to the climb, and the forces in the steady-state climb that follows.

@quiet flyer I have said repeatedly the closed vector diagram works.  The finer points are how "excess thrust" is apportioned.  At lower angles of climb we may be able to do really neat things like climb a little faster with lower AOA.

Now, for those who don't mind a little fun and controversy, let's flatten out pitch even more and and use excess lift, rather than excess thrust to accelerate upwards. What might this do to our trim?  Could we maybe need a little less (draggy) elevator downforce?  Is this starting to look a little like Vbg? Two observations to consider, a staticly nuetral plane will rise, but not pitch up strongly in a gust.  A rudder is defected the opposite way to hold line in a side slip.

So, we can see the "closed triangle" is a good starting point for the mechanistic modelling, but it will be tested.  (Optional evil laughter)

@RobertDiGiovanni -- see my comments under Sanchises' answer, they may clear up something that you are missing.

@quiet flyer ever hear of "cruise climb"?

First, we must be clear on what exactly is "excess thrust". I will list two possible definitions, although many more may be possible.

Excess thrust is the component of the resultant force in the direction of the flight path. Using this definition, excess thrust and steady state flight are directly at odds (because any net force results in an acceleration). So in that case, no, excess thrust is not possible in steady state flight.
Excess thrust is thrust minus drag. It splits the big lumpy thing called 'thrust' in two numbers: one that is used to counter drag (which is always there), and another that can be used for useful things.

Since we already answered the first option, I will now produce the free-body diagrams for the second option.

On the left, we have level flight. We recognize the four fundamental forces of flight: lift (blue), weight (green), drag (red), thrust (orange). It is immediately clear that all forces cancel each other, and we will happily coast along.
In the middle, we have a steady state climb. All vectors, except the weight vector, are 'tilted' back and change magnitude. Lift has decreased slightly, and as a consequence, drag has also decreased minutely (although this doesn't really matter). The biggest change can be seen in thrust, which has almost doubled in magnitude!
On the right image, I have proven that despite all these changes, the net force is still zero. I have replaced the lift- and drag vector in purple by subtracting drag from thrust. But wait a second! That purple vector is exactly what I called 'excess thrust': thrust minus drag! So this purple vector gives us a nice indication of how much 'extra' thrust we are using to make sure the resultant force on the airplane is still zero.
The sometimes heard term, "max excess thrust available" is then an indication of how far we can 'tilt back' the blue lift vector. On the diagram on the right, it should be obvious that this 'stretches' the purple 'excess thrust' vector. Steepest climb is thus achieved when the excess thrust equals the max excess thrust available.
Glossary

Resultant force: also called net force. The sum of all forces that act on the aircraft.
Steady state flight: all acceleration (in Earth's reference frame) is zero. I.e., level flight, steady climb, steady descent. Due to $\vec{F}=m\vec{a}$, zero acceleration means zero resultant force.

Addendum I: From the comments, "A Piper Cub cannot take part of its thrust and apply it for lift." We see in the image above that sadly, that is exactly what a Piper Cub must be doing to climb at all. The excess thrust $T_e$ requirement $T_e = W \sin \beta$ with $W$ lift and $\beta$ the climb angle. If very little excess thrust is available, we had better keep $\beta$ small - exactly what you would expect for a little cub!
Addendum II: below, the same image but with excess thrust according to definition 1. The vectors are no longer a 'closed triangle', and the net force is not zero. The aircraft is now accelerating.

Images created using IPE.

Comments are not for extended discussion; this conversation has been moved to chat.

When less than full thrust is in use during any steady state phase of flight, it can be considered that excess thrust is available.
Application of some or all of this excess thrust will result in a disturbance to the steady state - either acceleration in level flight, transition to a climb, increased climb rate, reduced descent rate or a combination, depending on what the pilot wants the airplane to do, eg at 10,000ft AMSL or FL100 accelerate from 250 kts and climb, once at climb speed, say 290 kts maintain the spd and continue climb at the resulting climb rate based on a fixed "climb thrust" setting throughout this sequence..
Just to clarify, there can be a steady state even during climb and descent if at constant spd and constant rate of climb and constant thrust, though in truth, this equation is also affected by the changes in properties described by the standard atmosphere.
Another way to look at this is that the calorific value of the fuel burnt by increasing thrust, puts energy into this system which is used for an increase in either kinetic energy (speed), or potential energy (altitude) or is shared by both.
At or close to the aircraft ceiling limitation, other factors come into play as also much excess thrust may just not be available, so that's another story.

Re :When less than full thrust is in use during any steady state phase of flight, it can be considered that excess thrust is available."-- isn't it pretty clear that that's not the sense of "excess thrust" intended by the original question?

Calorific value of the fuel burnt ... puts energy into the system, which is used for an increase in either (speed) or (altitude), yet they continue to show the thrust vector in their diagrams.  Its the effect of thrust (pouring energy in) and how the increased drag is used, increased speed or altitude (increased AOA).  Doesn't sound like you get less lift out of doing either one of those.

@skipper44 I liked your practical answer and the fact that you mentioned changes in the properties of the atmosphere.  I "fly my airspeed, not my throttle" because the planes aerodynamic qualities are dependent on speed.  It just concerns me a bit that folks are talking thrust, when excess thrust input may not be fast enough to increase V if the plane is stalling low and slow.  Excess thrust is converted into speed by F=ma.

@RobertDiGiovanni, Thrust comes from the engine, which in turn runs on fuel. To increase thrust you have to  increase fuel-flow. Achieving and maintaining a higher speed or altitude is entirely dependent upon the "calorific value of the fuel burnt by increasing thrust . . . "

That's true, I could also spill aviation fuel on the ground and light it, but that won't make the plane fly (lots of calorific value though). But now we're talking engine efficiency. Thrust can only accelerate the aircraft.  So, if one is too slow, and can't accelerate fast enough, they are $Screwed$.

Exactly my point, unless the engine burns that extra fuel, you would be constrained to sharing the existing energy of the system, ie trade (drop) speed to gain Altitude, or trade (lose) altitude to gain speed. We're talking about energy share, and the Law of Conservation of Energy, Efficiency is v much part of this equation, but not directly  relevant to this discussion.

In a steady-state climb, Thrust is not a leg of the closed triangle of force vectors; rather, (Thrust-Drag) is.
See the right-hand vector diagram in this related answer-- Is excess lift or excess power needed for a climb?.  The diagram shows that if our definition of "excess thrust" is (Thrust minus Drag), then excess Thrust clearly does exist in a steady-state climb.
The present question seems to imply that if we are flying horizontally, and then we increase Thrust, we must increase angle-of-attack if we wish to climb rather than accelerate horizontally.  This isn't actually the case, at least over the long run.  If we hold angle-of-attack constant and increase Thrust, we'll have a brief initial acceleration, but we'll end up in a steady-state climb at a slightly lower airspeed than we started with, at least in the case where little or no downthrust is present in relation to the actual direction of the flight path at any instant.
The reason for this is that for any given angle-of-attack, the lift and drag coefficients are constant, so airspeed scales according to the square root of the Lift vector, which also means that airspeed scales according to the square root of the Drag vector.  And we know that Lift is slightly less in a steady-state climb than in wings-level flight, according to the relationship Lift = Weight * cosine (climb angle), at least in the case where no downthrust or upthrust is present relative to the direction of the flight path.
Note the relationship between Thrust and airspeed on either of the two right-hand tables in this related answer-- Is excess lift or excess power needed for a climb?.
The right-hand tables assume that no upthrust or downthrust is present, relative to the direction of the flight path.  Each table assumes a constant angle-of-attack from one row to the next.
(But the airspeed values are not intended to be scaled in the same units between the two tables, since the lift coefficient is not specified, and is presumably much less in the upper table than in the lower table.)
You may find the vector diagrams, calculations, and explanations in this other related answer to be helpful as well -- Does lift equal weight in a climb?